CF Round946 (Div. 3)C：容斥

Beautiful Triple Pairs

题目描述

Polycarp was given an array $ a $ of $ n $ integers. He really likes triples of numbers, so for each $ j $ ( $ 1 \le j \le n - 2 $ ) he wrote down a triple of elements $ [a_j, a_{j + 1}, a_{j + 2}] $ .

Polycarp considers a pair of triples $ b $ and $ c $ beautiful if they differ in exactly one position, that is, one of the following conditions is satisfied:

• $ b_1 \ne c_1 $ and $ b_2 = c_2 $ and $ b_3 = c_3 $ ;
• $ b_1 = c_1 $ and $ b_2 \ne c_2 $ and $ b_3 = c_3 $ ;
• $ b_1 = c_1 $ and $ b_2 = c_2 $ and $ b_3 \ne c_3 $ .

Find the number of beautiful pairs of triples among the written triples $ [a_j, a_{j + 1}, a_{j + 2}] $ .

输入格式

The first line contains a single integer $ t $ ( $ 1 \le t \le 10^4 $ ) — the number of test cases.

The first line of each test case contains a single integer $ n $ ( $ 3 \le n \le 2 \cdot 10^5 $ ) — the length of the array $ a $ .

The second line of each test case contains $ n $ integers $ a_1, a_2, \dots, a_n $ ( $ 1 \le a_i \le 10^6 $ ) — the elements of the array.

It is guaranteed that the sum of the values of $ n $ for all test cases in the test does not exceed $ 2 \cdot 10^5 $ .

输出格式

For each test case, output a single integer — the number of beautiful pairs of triples among the pairs of the form $ [a_j, a_{j + 1}, a_{j + 2}] $ .

Note that the answer may not fit into 32-bit data types.

样例 #1

样例输入 #1

8
5
3 2 2 2 3
5
1 2 1 2 1
8
1 2 3 2 2 3 4 2
4
2 1 1 1
8
2 1 1 2 1 1 1 1
7
2 1 1 1 1 1 1
6
2 1 1 1 1 1
5
2 1 1 1 1

样例输出 #1

2
0
3
1
8
4
3
2

提示

In the first example, $ a = [3, 2, 2, 2, 3] $ , Polycarp will write the following triples:

1. $ [3, 2, 2] $ ;
2. $ [2, 2, 2] $ ;
3. $ [2, 2, 3] $ .

The beautiful pairs are triple $ 1 $ with triple $ 2 $ and triple $ 2 $ with triple $ 3 $ .In the third example, $ a = [1, 2, 3, 2, 2, 3, 4, 2] $ , Polycarp will write the following triples:

1. $ [1, 2, 3] $ ;
2. $ [2, 3, 2] $ ;
3. $ [3, 2, 2] $ ;
4. $ [2, 2, 3] $ ;
5. $ [2, 3, 4] $ ;
6. $ [3, 4, 2] $ ;

The beautiful pairs are triple $ 1 $ with triple $ 4 $ , triple $ 2 $ with triple $ 5 $ , and triple $ 3 $ with triple $ 6 $ .

AC_code

要统计一位不同，即两位相同的次数，但要注意当完全一样即会被计算3次，要减去3次

#include <bits/stdc++.h>

using namespace std;

using LL = long long;

const int N = 2 * 1e5 + 10;

int a[N];
int _, n;

void solve()
{
    cin >> n;
    for(int i = 1; i <= n; ++ i) cin >> a[i];
    LL ans = 0;
    
    map<pair<int, int>, int> mp1, mp2, mp3;
    map<vector<int>, int> cnt;
    
    for(int i = 1; i + 2 <= n; ++ i) {
        ans += mp1[{a[i], a[i + 1]}] ++;
        ans += mp2[{a[i], a[i + 2]}] ++;
        ans += mp3[{a[i + 1], a[i + 2]}] ++;
        
        vector<int> v = {a[i], a[i + 1], a[i + 2]};
        ans -= cnt[v] * 3;
        cnt[v] ++;
    }
    cout << ans << endl;
}

int main()
{
    cin >> _;
    while(_ -- ) {
        solve();
    }
    return 0;
}

利用偏移量构造2个相等

学到了，不知道下次实战能不能写出来QAQ

#include <bits/stdc++.h>

using namespace std;

using i64 = long long;

int n;
void solve() {
    cin >> n;
    vector<int> a(n);
    for(int i = 0; i < n; ++ i) cin >> a[i];

    i64 ans = 0;
    //清楚的作用域让你思路更清楚
    for(int i = 0; i < 3; ++ i) {//这里利用偏移量i=0,j=1,2和i=1,j=2对应3个关系
        for(int j = i + 1; j < 3; ++ j) {
            cout << "j = " << j << endl;
            map<pair<int, int>, int> cnt;
            for(int k = 0; k + 2 < n; ++ k) {
                ans += cnt[{a[k + i], a[k + j]}] ++;
            }
        }
    }
    
    map<tuple<int, int, int>, int> cnt;//元组
    for(int i = 0; i + 2 < n; ++ i) {
        ans -= 3 * cnt[{a[i], a[i + 1], a[i + 2]}] ++;//还是全相等即减去
    }
    cout << ans << "\n";
}

void init() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
}

int _;
int main()
{
    init();
    cin >> _;

    while(_ --) {
        solve();
    }
    return 0;
}

本人思路

一眼样例感觉好像是全排列+处理冗余，但是需要额外处理每个数能用多少次(这个显然我不会)，继续看样例发现是按长度为3依次枚举，转而想用最近学到的那个分段枚举，但是不同1个的串的顺序不同可能导致问题，最后直接暴力扫一遍维护长度为3的串匹配

//错误代码
int res = 0;
    int v[3] = {0};
    int r[3] = {0};
    for(int i = 1; i <= n - 2; ++ i) {
        v[0] = q[i], v[1] = q[i + 1], v[2] = q[i + 2];
        for(int j = i + 1; j < i + 4; ++ j) {
                c = 0;
                if(j + 2 > n) break;
                r[0] = q[j], r[1] = q[j + 1], r[2] = q[j + 2];
                for(int i1 = 0; i1 < 3; ++ i1)
                    for(int j1 = 0; j1 < 3; ++ j1)
                        if(v[i1] == r[j1]) ++ c;
                if(c >= 2) ++ res;
        }
    }